Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 9x}{x - 8} = \dfrac{21x - 20}{x - 8}$
Answer: Multiply both sides by $x - 8$ $ \dfrac{x^2 + 9x}{x - 8} (x - 8) = \dfrac{21x - 20}{x - 8} (x - 8)$ $ x^2 + 9x = 21x - 20$ Subtract $21x - 20$ from both sides: $ x^2 + 9x - (21x - 20) = 21x - 20 - (21x - 20)$ $ x^2 + 9x - 21x + 20 = 0$ $ x^2 - 12x + 20 = 0$ Factor the expression: $ (x - 2)(x - 10) = 0$ Therefore $x = 2$ or $x = 10$ The original expression is defined at $x = 2$ and $x = 10$, so there are no extraneous solutions.